this should work just fine if you just need to show what time left, since JavaScript uses frames for its time you'll have get your End Time - The Time RN after that we can divide it by 1000 since apparently 1000 frames = 1 seconds, after that you can use the basic math of time, but there's still a problem to this code, since the calculation is static, it can't compensate for the different day total in a year (360/365/366), the bunch of IF after the calculation is to make it null if the time is lower than 0, hope this helps even though it's not exactly what you're asking :)

var now = new Date();
var end = new Date("End Time");
var total = (end - now) ;
var totalD =  Math.abs(Math.floor(total/1000));

var years = Math.floor(totalD / (365*60*60*24));
var months = Math.floor((totalD - years*365*60*60*24) / (30*60*60*24));
var days = Math.floor((totalD - years*365*60*60*24 - months*30*60*60*24)/ (60*60*24));
var hours = Math.floor((totalD - years*365*60*60*24 - months*30*60*60*24 - days*60*60*24)/ (60*60));
var minutes = Math.floor((totalD - years*365*60*60*24 - months*30*60*60*24 - days*60*60*24 - hours*60*60)/ (60));
var seconds = Math.floor(totalD - years*365*60*60*24 - months*30*60*60*24 - days*60*60*24 - hours*60*60 - minutes*60);

var Y = years < 1 ? "" : years + " Years ";
var M = months < 1 ? "" : months + " Months ";
var D = days < 1 ? "" : days + " Days ";
var H = hours < 1 ? "" : hours + " Hours ";
var I = minutes < 1 ? "" : minutes + " Minutes ";
var S = seconds < 1 ? "" : seconds + " Seconds ";
var A = years == 0 && months == 0 && days == 0 && hours == 0 && minutes == 0 && seconds == 0 ? "Sending" : " Remaining";

document.getElementById('txt').innerHTML = Y + M + D + H + I + S + A;

另一种解决方案是将difference转换为一个新的Date对象，并获得该日期的年(与1970年不同)、月、日等。

var date1 = new Date(2010, 6, 17);
var date2 = new Date(2013, 12, 18);
var diff = new Date(date2.getTime() - date1.getTime());
// diff is: Thu Jul 05 1973 04:00:00 GMT+0300 (EEST)

console.log(diff.getUTCFullYear() - 1970); // Gives difference as year
// 3

console.log(diff.getUTCMonth()); // Gives month count of difference
// 6

console.log(diff.getUTCDate() - 1); // Gives day count of difference
// 4

所以差异就像“3年6个月零4天”。如果你想以一种人类可读的风格呈现不同，这将对你有所帮助。

如果你使用的是moment.js，那么查找日期差异就很简单了。

var now  = "04/09/2013 15:00:00";
var then = "04/09/2013 14:20:30";

moment.utc(moment(now,"DD/MM/YYYY HH:mm:ss").diff(moment(then,"DD/MM/YYYY HH:mm:ss"))).format("HH:mm:ss")

function DateDiff(b, e)
{
    let
        endYear = e.getFullYear(),
        endMonth = e.getMonth(),
        years = endYear - b.getFullYear(),
        months = endMonth - b.getMonth(),
        days = e.getDate() - b.getDate();
    if (months < 0)
    {
        years--;
        months += 12;
    }
    if (days < 0)
    {
        months--;
        days += new Date(endYear, endMonth, 0).getDate();
    }
    return [years, months, days];
}

[years, months, days] = DateDiff(
    new Date("October 21, 1980"),
    new Date("July 11, 2017")); // 36 8 20

抱歉，平毫秒计算不可靠 感谢所有的回复，但我尝试的功能很少失败 1. 一个接近今天的日期 2. 1970年或者 3.闰年的一天。

最适合我的方法，涵盖了所有场景，例如闰年，接近1970年的日期，2月29日等。

var someday = new Date("8/1/1985");
var today = new Date();
var years = today.getFullYear() - someday.getFullYear();

// Reset someday to the current year.
someday.setFullYear(today.getFullYear());

// Depending on when that day falls for this year, subtract 1.
if (today < someday)
{
    years--;
}
document.write("Its been " + years + " full years.");

假设你有两个Date对象，你可以减去它们，以毫秒为单位得到差值:

var difference = date2 - date1;

从那里，您可以使用简单的算术来推导其他值。